Physics: What are the x- and y- components of a vector that has a 42.9 m magnitude and a direction of 252

I think its labor but not 100 percent ShureExplanation:...

What are the x- and y- components of a vector that

arielpraczko1

05.01.2022

-13.3 m, -40.8 m

Explanation:

We can find the components of the vector by using the equations:

$v_x = v cos \theta$

$v_y = v sin \theta$

where

v is the magnitude of the vector

$\theta$ is the angle representing the direction of the vector (measured above the positive x-direction)

For the vector in this problem,

v = 42.9 m

$\theta=252^{\circ}$

Therefore its components are

$v_x = (42.9) ( cos 252^{\circ})=-13.3 m\\v_y = (42.9)( sin 252^{\circ})=-40.8 m$

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